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3 years ago

What are the factor that affect the efficiency of a pulley

Physics

1 answer:

alukav5142 [94]3 years ago

8 0

Answer:

Tension in the chains - In a chain drive, technically, you have a closed-chain (which has no end) going around 2 pulley or gears; looking closely you have 2 parallel chains going in opposite direction. If kept in horizontal direction, the one below the other is the slack side and the other the tight side. The tension on the upper or tight side is more than the slack side. So you need to keep in mind to keep your chain drive tight so that there is no loss or rotation or lags.

Sizes of the pulley/gear - The chain will be warped around a pair of pulley or gear. The sizes of these pulley/gear will also determine the efficiency of the chain drive (consider one big and one small)

Number of pulley/gear - If the number of pulley/gear is more and chain wrapped on it with little complexity will result in decrease in efficiency because of extra tension.

Length of the chain drive - You cannot have much too long chain drive. It will make your slack side more heavy because the end are further away. You have to apply more power and possibilities of lag increases decreasing efficiency. In an ideal situation, this won't happen, but this world isn't ideal.

Friction between chains & pulley/gear - If you have studied gears (involving its teeth), you will come to know that there is friction offered on the two meeting surfaces.

Angle of contact - This would have been explained better with a diagram. Although, if you are familiar with the terms you won't have difficulty understanding. Angle of contact is the angle the chain forms with the pulley/gear at the point of contact with the center of the pulley. The angle of contact should not be too small, or else the things will be slippery.

Explanation:

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2 years ago

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Answer:

h'=0.25m/s

Explanation:

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$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

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Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

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What are the factor that affect the efficiency of a pulley​

What are the factor that affect the efficiency of a pulley